The density of material at any point in a solid cylinder, y^2 + z^2 ≤ a^2, 0 ≤ x ≤ b is proportional to the distance of the point from the x-axis. Find the density. Use k as the proportionality constant. δ(y, z) = δ(r) =
I think I got the density
\[k \sqrt{y^2+z^2}\]
what do you think @wio ?
It would help if you draw it.
But I believe that your density function is correct
how would I change it to polar coordinates? @wio
\[ r^2 = y^2+z^2 \]
\[ y=r\sin\theta,\, z=r\cos\theta \]
And \(x\) remains the same as the height component
Since it is a cylinder \(0\leq \theta\lt 2\pi\). Since \(x\) acts as the height, it can remain the same.
thanks @wio :)
wait so I don't haven an x on my density funtion @wio
\[k \sqrt{rsin \theta+rcos \theta} \]
No, you don't, because the \(x\) value doesn't change the density
Changing the \(x\) value doesn't change the distance from the \(x\) axis
\[ \delta (y,z) = k\sqrt{y^2+z^2} = k\sqrt{r^2} = kr \]
Well \(k\sqrt{r^2} = k|r| = kr\), we can say \(|r|=r\) because by design \(r\geq0\).
I don't like how they have \(\delta (y,z)\) and \(\delta (r)\). They are two different functions. I suppose that is me nitpicking though.
For example \(\delta(r,r) = k\sqrt{r^2+r^2} = kr\sqrt 2\neq kr=\delta(r)\).
for the second part of this question they ask to set up an integral
\[\int\limits\limits_{?}^{?}\int\limits\limits_{?}^{?}\int\limits\limits_{?}^{?}f(x,r,\theta)dx dr d \theta \]
so it would be \[\int\limits_{0}^{2\pi}\int\limits_{0}^{6}\int\limits_{0}^{5}kr dx d r d \theta \]
cylinder is radius 6 and length 5 @wio
Remember the Jacobian
\[ dydz =rdrd\theta \]
what's the jacobian?
It's the scaling factor of the differentials
so it would be \[dx r dr d \theta \]
what whould the k value be?
proportinality constant
they ask for exact answer by having k I can not find an exact answer
You can get an exact answer in terms of \(k\)
can k just be x
No, because that wouldn't work
\(k\) is constant with respect to \(x,y,z\)
i'm so confuse I'm sorry
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